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In Problems 1-6 write the given nonlinear second-order differential equation as a plane autonomous system. Find all critical points of the resulting system. $$ x^{\prime \prime}+x-\epsilon x|x|=0 \text { for } \epsilon>0 $$

Short Answer

Expert verified
The critical points are \((0,0)\) and \((\frac{1}{\epsilon}, 0)\).

Step by step solution

01

Rewrite the second-order equation as a system of first-order equations

Given the second-order differential equation: \[ x'' + x - \epsilon x|x| = 0 \] Introduce a substitution by letting \( x' = y \), which implies that \( x'' = y' \). Now, rewriting the initial equation: \[ y' = -x + \epsilon x|x| \]So we can write the system of first-order equations as: \[ \begin{align*} x' &= y \ y' &= -x + \epsilon x|x| \end{align*} \]
02

Identify the autonomous system

The resulting system: \[ \begin{align*} x' &= y \ y' &= -x + \epsilon x|x| \end{align*} \] This is a plane autonomous system because both functions on the right-hand side do not explicitly depend on time.
03

Find the critical points

A critical point for the system occurs where both derivatives \( x' \) and \( y' \) are zero. This leads to the equations:1. \( x' = y = 0 \)2. \( y' = -x + \epsilon x|x| = 0 \)From the first equation, \( y = 0 \). Substituting into the second equation:\[ -x + \epsilon x|x| = 0 \]Simplifying gives:\[ \epsilon x^2 = x \]If \( x eq 0 \), divide both sides by \( x \):\[ \epsilon x = 1 \Rightarrow x = \frac{1}{\epsilon} \]Checking \( x = 0 \): The second equation is satisfied at \( x = 0 \) as well. Thus, the critical points are \( (0,0) \) and \( (\frac{1}{\epsilon}, 0) \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Plane Autonomous System
An autonomous system is a set of equations where the derivatives of the variables are functions of the variables themselves, not explicitly involving another independent variable like time. In our original problem with the differential equation \( x^{\prime \prime}+x-\epsilon x|x|=0 \), we convert it to a first-order system by introducing the variable \( y \) such that \( x' = y \). This approach results in:
  • \( x' = y \)
  • \( y' = -x + \epsilon x|x| \)
This system does not explicitly involve time, hence it is labeled a plane autonomous system. Here, 'plane' denotes the two-dimensional space created by the variables \( x \) and \( y \). Autonomous systems are useful for analyzing stability and behavior of systems through phase portraits, without considering external time factors.
Critical Points
Critical points in a system occur where the rate of change of all variables is zero. For our system:
  • \( x' = y = 0 \)
  • \( y' = -x + \epsilon x|x| = 0 \)
Setting \( x' = y = 0 \) simplifies finding critical points. With \( y = 0 \), substitution into \( y' = -x + \epsilon x |x| = 0 \) leads us to solve \( -x + \epsilon x^2 = 0 \). Factoring and solving gives the critical points:
  • \( (0, 0) \)
  • \( \left(\frac{1}{\epsilon}, 0\right) \)
These points indicate where the system is in equilibrium, meaning from these points, the system would not change state if undisturbed. Analyzing critical points helps in understanding the long-term behavior and stability of a dynamical system.
Second-order Differential Equation
A second-order differential equation involves the second derivative of a function. In the exercise, the equation \( x^{\prime \prime} + x - \epsilon x|x| = 0 \) is of second order because \( x'' \) appears in it. Typically, such equations represent systems where acceleration or curvature is involved, like harmonic oscillators. To solve or analyze them, it is often helpful to reduce the order of the equation by expressing it as a system of first-order equations, which simplifies the complexity and allows for more straightforward analysis. Converting into first-order systems is a powerful method because it often reveals the underlying structure and symmetry of the system through matrices or geometric considerations.
First-order System of Equations
A first-order system of equations is comprised of equations with only first derivatives. In the solution provided, we transformed the second-order equation into this type by using a substitution:
  • Introduce \( y = x' \)
  • Write \( x'' \) as \( y' \)
The resulting system:
  • \( x' = y \)
  • \( y' = -x + \epsilon x|x| \)
Such systems are advantageous because they allow for the application of linear algebra techniques and visualization through phase plane analysis. The dimensionality reduction from higher-order derivatives to first-order makes assessing behavior over time more approachable and is particularly useful for finding critical points and evaluating system stability. This transformation underlies many modeling techniques in physics and engineering.

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Most popular questions from this chapter

Answer Problems 1-10 without referring back to the text. Fill in the blank or answer true/false. All solutions to the pendulum equation \(\frac{d^{2} \theta}{d t^{2}}+\frac{g}{l} \sin \theta=0\) are periodic.

Without solving explicitly, classify (if possible) the critical points of the autonomous first-order differential equation \(x^{\prime}=\left(x^{2}-1\right) e^{-2 / 2}\) as asymptotically stable or unstable.

Use the phase-plane method to show that the solutions to the nonlinear second- order differential equation \(x^{\prime \prime}=-2 x \sqrt{\left(x^{\prime}\right)^{2}+1}\) that satisfy \(x(0)=x_{0}\) and \(x^{\prime}(0)=0\) are periodic.

If we assume that a damping force acts in a direction opposite to the motion of a pendulum and with a magnitude directly proportional to the angular velocity \(d \theta / d t\), the displacement angle \(\theta\) for the pendulum satisfies the nonlinear second-order differential equation $$ m l \frac{d^{2} \theta}{d t^{2}}=-m g \sin \theta-\beta \frac{d \theta}{d t} . $$ (a) Write the second-order differential equation as a plane autonomous system, and find all critical points. (b) Find a condition on \(m, l\), and \(\beta\) that will make \((0,0)\) a stable spiral point.

In Problems \(23-26\) solve the given nonlinear plane autonomous system by changing to polar coordinates. Describe the geometric behavior of the solution that satisfies the given initial condition(s). \(x^{\prime}=-y-x\left(x^{2}+y^{2}\right)^{2}\) \(y^{\prime}=x-y\left(x^{2}+y^{2}\right)^{2}, \quad \mathbf{X}(0)=(4,0)\)

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